Chapter 11
Solutions
The Composition of Solutions
The Solution Process
Factors that Affect Solubility
Measuring Concentrations of Solutions
Quantities for Reactions That Occur in
Aqueous Solution
• Colligative Solutions
•
•
•
•
•
11-1
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Composition of Solutions
• A solution is:
– A homogeneous mixture with uniform
composition throughout
– Composed of:
• Solute
– Substance being dissolved
– Usually present in the lesser amount
• Solvent
– Substance doing the dissolving
– Usually present in the greater amount
11- 2
Composition of Solutions
• What is the solute? What is the solvent?
11- 3
1
Composition of Solutions
Figure 11.5
11- 4
Solubility
• How readily or completely a solute
dissolves in a solvent
– A soluble ionic compound dissociates into
ions when dissolved in water.
– Insoluble compounds remain mostly
undissociated in their ionic, crystalline
structure.
• The solubilities for ionic compounds do
not follow a predictable pattern
– Must be recalled from experimental data
– Solubility rules are listed in Table 11.1
11- 5
Rules Used to Predict the Solubility of Ionic Salts
Ions
Rules
Na+, K+, NH4+
Most salts of sodium, potassium, and ammonium ions are soluble.
NO3-
All nitrates are soluble.
SO42-
Most sulfates are soluble. Exceptions: BaSO4, SrSO4, PbSO4,
CaSO4, Hg2SO4, & Ag2SO4
Cl-, Br-, I-
Most chlorides, bromides, and iodides are soluble. Exceptions:
AgX, Hg2X, PbX2 , & HgI2 (X = Cl, Br, or I)
Ag+
Silver salts, except AgNO3, are insoluble.
O2-, OH-
Oxides and hydroxides are insoluble. Exceptions: NaOH, KOH,
Ba(OH)2, & Ca(OH)2 (somewhat soluble)
S2-
Sulfides are insoluble. Exceptions: Salts of Na+, K+, NH4+, and the
alkaline earth metal ions
CrO42-
Most chromates are insoluble. Exceptions: Salts of Na+, K+, NH4+,
Mg2+, Ca2+, Al3+, and Ni2+
CO32-, PO43-, SO32-,
SiO32-
Most carbonates, phosphates, sulfites, and silicates are insoluble.
Exceptions: Na+, K+, and NH4+
11- 6
2
Electrolytes
• A solution that conducts electricity
• Two kinds of electrolytes:
– Strong electrolyte
• A solute that dissociates completely into
ions in aqueous solution
– Weak electrolyte
• A solute that dissociates partially into ions
in aqueous solution
• Nonelectrolyte
– A solute that does not dissociate into
ions in aqueous solution
11- 7
Electrolytes
11- 8
Practice – Composition of
Solutions
• What ions, atoms, or molecules are
present when the following
substances are mixed with water?
Write a balanced equation to
represent the process of dissolving
each substance in water.
– ZnCl2(s)
– KNO3(s)
– C5H11OH(l)
11- 9
3
Practice Solutions – Composition
of Solutions
– ZnCl2(s)
ZnCl2(s) + H2O(l) Zn2+(aq) + 2 Cl-(aq)
– KNO3(s)
KNO3(s) + H2O(l) K+(aq) + NO3-(aq)
– C5H11OH(l)
C5H11OH(l) + H2O(l) C5H11O-(aq) + H+(aq)
Does not dissolve extensively…
11-10
Like Dissolves Like
• When the bonding in a solvent is similar to the bonding in a
solute, then the solute will dissolve in the solvent.
– Polar covalent solvents dissolve polar covalent solutes.
– Nonpolar covalent solvents dissolve nonpolar covalent
solutes.
– See Table 11.2 below:
Solute
Solvent
Polar
Nonpolar
Ionic
Soluble
Insoluble
Polar
Soluble
Insoluble
Nonpolar Insoluble
Soluble
11-11
Solution Process
Figure 11.12
11-12
4
Solution Process
•
Ion-dipole force
– An attraction between an
ion and a polar molecule
•
When an ionic
compound dissolves in
water:
– Ionic bonds in the solute
break.
– Hydrogen bonds between
water molecules break.
– Ion-dipole forces form
between ions and water
molecules.
11-13
Solution Process
11-14
Heat of Solution
• The difference between the energy
required for separation of solute and
solvent and the energy released upon
formation of ion-dipole interactions
• When the solvation process provides
more energy than is needed to separate
the pure solvent and solvent particles, the
heat of solution is negative, and the
overall dissolving process is exothermic.
11-15
5
Endothermic Solution Process
Figure 11.14B
11-16
Solution Process for NonpolarNonpolar Interactions
11-17
Entropy
• A measure of the tendency for
matter to become disordered or
random in its distribution
• Matter spontaneously changes from
a state of order to a state of
randomness, unless energy changes
oppose it.
• Solutions form because the makeup
of the solution naturally tends
toward disorganization.
11-18
6
Entropy
Figure 11.16
11-19
Practice – The Solution Process
• Iodine (I2) dissolves in hexane
(C6H14) to form a solution. Describe
the forces that must be broken for
the solution to form.
11- 20
Practice Solutions – The Solution
Process
• Iodine (I2) dissolves in hexane
(C6H14) to form a solution. Describe
the forces that must be broken for
the solution to form.
London forces must be broken to
allow I2, the solute, to dissolve.
London forces must also be broken
to allow C6H14, the solvent, to
dissolve.
11- 21
7
Factors that Affect Solubility
• Three factors
affect
solubility:
– Structure
– Temperature
– Pressure
11- 22
Structure
•
For a solid to be soluble in a
given solvent, the energy
released by the solvation
process must compensate
for all of the energy required
to break up the forces at
work both within the solute
and the solvent.
–
•
Entropy is also a factor
The solubility (or miscibility)
of liquids in other liquids
depends largely on the
polarity of the solute and
solvent molecules.
11- 23
Temperature
•
Affects the solubility of most substances in
water
– Most ionic solids are more soluble in water at higher
temperatures
– Gases are less soluble as temperature increases
Figure 11.19
11- 24
8
Pressure
• Affects gases but not liquids or
solids
• The solubility of a gas in a liquid is
directly proportional to the pressure
of the gas above the liquid
• An increase in pressure results in an
increase in solubility of a gas
11- 25
Pressure
Figure 11.20
11- 26
Measuring Concentrations of Solutions
•
Concentration
– The relative amounts of the solutes and solvent that
make up the solution
Concentration =
•
amount of solute
amount of solution (or solvent)
Solubility
– The ratio that identifies the maximum amount of a
solute that will dissolve in a particular solvent to
form a stable solution under specified conditions
– A measure of maximum concentration
– Three types of concentrated solution:
• Saturated solution
• Unsaturated solution
• Supersaturated solution
11- 27
9
Saturated Solution
•
A solution that contains the
maximum possible amount
of solute
– When this type of
solution forms, a
dynamic equilibrium is
established.
• Undissolved solute
continues to dissolve,
but at the same time,
previously dissolved
solute is deposited
from solution.
11- 28
Supersaturated Solution
• A solution that contains more than the
maximum possible amount of solute
• Unsaturated solution
– A solution that contains less than the
maximum possible amount of solute
Figure 11.23
11- 29
Table 11.3: Concentration Units
Unit
Definition
Percent by mass
grams of solute x 100%
grams of solution
Percent by volume
volume of solute x 100%
volume of solution
Mass/volume percent
grams of solute x 100%
volume of solution
Parts per million
grams of solute x 106
grams of solution
Parts per billion
grams of solute x 109
grams of solution
Molarity (M)
moles of solute
liters of solution
Molality (m)
moles of solute
kilograms of solvent
11- 30
10
Percent by Mass
• Is calculated by dividing the solute
mass by the mass of the solution
and multiplying it by 100%
Percent by mass = mass (solute) x 100%
mass (solution)
• To find the mass of the solution, we
can add the mass of the solute to the
mass of the solvent.
11- 31
Percent by Mass
•
Example:
– What is the percent-by-mass concentration of
acetic acid (CH3CO2H) in a vinegar solution
that contains 2.70 g of acetic acid and 122.8 g
of water?
Percent by mass = mass (solute) x 100%
mass (solution)
Percent by mass =
2.70 g
x 100%
(2.70 g + 122.8 g)
Percent by mass = 2.15%
11- 32
Percent by Volume
• Is calculated by dividing the solute
volume by the solution volume and
multiplying by 100%
Percent by volume = volume (solute) x 100%
volume (solution)
• The volume of the solution must be
measured. Volumes are not additive
like masses.
11- 33
11
Percent by Volume
•
Example:
– To prepare a solution, we dissolve 205 mL of ethanol
in sufficient water to give a total volume of 235 mL.
What is the percent by volume concentration of
ethanol?
Percent by volume = volume (solute) x 100%
volume (solution)
Percent by volume = 205 mL x 100%
235 mL
Percent by volume = 87.2%
11- 34
Mass/Volume Percent
• Is calculated by dividing the solute mass
by the volume of solution and multiplying
by 100%
Mass/Volume Percent = grams of solute x 100%
volume of solution
• Example:
– The concentration of NaCl in a saline bag is
0.9%. Therefore, for every 100 mL of
solution, the mass of NaCl present is 0.9 g.
11- 35
Parts per Million and per Billion
• Parts per million
– Is calculated by dividing the mass of solute
by the mass of solution and multiplying by 1
million (106)
Parts per million = mass (solute) x 106
mass (solution)
• Parts per billion
– Is calculated by dividing the mass of solute
by the mass of solution and multiplying by 1
billion (109)
Parts per billion = mass (solute) x 109
mass (solution)
11- 36
12
Molarity
• The ratio of moles of solute to the volume
of solution in liters
Molarity (M) = moles (solute)
liters (solution)
• Example
– A solution contains 117 g of potassium
hydroxide (KOH) dissolved in sufficient water
to give a total volume of 2.00 L. The molar
mass of KOH is 56.11 g/mol. What is the
molarity of the aqueous potassium hydroxide
solution?
11- 37
Molarity
•
Example
– A solution contains 117 g of potassium hydroxide (KOH)
dissolved in sufficient water to give a total volume of 2.00
L. The molar mass of KOH is 56.11 g/mol. What is the
molarity of the aqueous potassium hydroxide solution?
Molarity (M) = moles (solute)
liters (solution)
Moles KOH = 117 g KOH x 1 mol KOH = 2.09 mol KOH
56.11 g KOH
Molarity (M) = 2.09 mol KOH = 1.04 M KOH
2.00 L
11- 38
Molality
• If the temperature of a solution increases,
the molarity of the solution changes
because the volume changes.
• Molality does not change with
temperature because it divides moles of
solute with the mass of solution.
molality (m) =
moles (solute)
kilograms (solution)
11- 39
13
Molality
•
Example
– A solution contains 56.7 g of potassium hydroxide (KOH)
dissolved in sufficient water to give a total mass of 245.8
g. The molar mass of KOH is 56.11 g/mol. What is the
molality of the aqueous potassium hydroxide solution?
molality (m) =
moles (solute)
kilograms (solution)
moles (solute) = 56.7 g KOH x 1 mol KOH = 1.01 mol KOH
56.11 g KOH
kilograms (solvent) = 245.8 g x 1 kg = 0.2458 kg
1000 g
11- 40
Molality
•
Example
– A solution contains 56.7 g of potassium
hydroxide (KOH) dissolved in sufficient water to
give a total mass of 245.8 g. The molar mass of
KOH is 56.11 g/mol. What is the molality of the
aqueous potassium hydroxide solution?
molality (m) =
moles (solute)
kilograms (solution)
molality (m) = 1.01 mol KOH = 4.11 m KOH
0.2458 kg KOH
11- 41
Precipitation Reactions
• The cation from one reactant combines
with the anion from another reactant to
form an insoluble compound, called a
precipitate.
• Example
– When a solution of lead(II) nitrate is mixed
with a solution of potassium chromate, a
yellow precipitate forms according to the
equation:
Pb(NO3)2(aq) + K2CrO4(aq) 2 KNO3(aq) + PbCrO4(s)
What volume of 0.105 M lead(II) nitrate is
required to react with 100.0 mL of 0.120 M
potassium chromate? What mass of PbCrO4
solid forms?
11- 42
14
Precipitation Reactions
Pb(NO3)2(aq) + K2CrO4(aq) 2 KNO3(aq) + PbCrO4(s)
What volume of 0.105 M lead(II) nitrate is required to
react with 100.0 mL of 0.120 M potassium chromate?
What mass of PbCrO4 solid forms?
To solve for volume of Pb(NO3)2:
M x V = moles
100.0mL K2CrO4 x 0.120 moles K2CrO4 = 0.0120 moles K2CrO4
1000 mL K2CrO4
0.0120 moles K2CrO4 x 1 mol Pb(NO3)2 = 0.0120 moles Pb(NO3)2
1 mol K2CrO4
0.0120 moles Pb(NO3)2 x 1 L Pb(NO3)2
= 0.114 L Pb(NO3)2
0.105 mol Pb(NO3)2
11- 43
Precipitation Reactions
Pb(NO3)2(aq) + K2CrO4(aq) 2 KNO3(aq) + PbCrO4(s)
What volume of 0.105 M lead(II) nitrate is required to
react with 100.0 mL of 0.120 M potassium chromate?
What mass of PbCrO4 solid forms?
To solve for volume of Pb(NO3)2:
M x V = moles
0.0120 moles K2CrO4 * 1 mol PbCrO4 = 0.0120 moles PbCrO4
1 mol K2CrO4
0.0120 moles PbCrO4 * 323.2 g PbCrO4 = 3.88 g PbCrO4
1 mol PbCrO4
11- 44
Acid-Base Titrations
• Titration
– The process of determining the concentration of one
substance in solution by reacting it with a solution of
another substance that has a known concentration
• Equivalence point
– The point in a titration at which the moles of the acid
equal the moles of the base
• End point
– The point in a titration where the indicator changes
color
– Slightly different than the equivalence point, but a
good estimation
11- 45
15
Acid-Base Titrations
11- 46
Practice - Acid-Base
Titrations
• We need 28.18 mL of 0.2437 M HCl
solution to completely titrate 25.00 mL of
Ba(OH)2 solution of unknown
concentration. What is the molarity of the
barium hydroxide solution?
11- 47
Practice Solutions - AcidBase Titrations
• We need 28.18 mL of 0.2437 M HCl solution to completely
titrate 25.00 mL of Ba(OH)2 solution of unknown
concentration. What is the molarity of the barium
hydroxide solution?
First, write the balanced chemical
equation:
2 HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2 H2O(l)
Next, find the moles of HCl:
28.17 mL HCl x
0.2437 moles HCl
= 6.865 x 10-3 moles HCl
1000 mL HCl
11- 48
16
Practice Solutions - Acid-Base
Titrations
HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + H2O(l)
Next, we need to find the volume (in L) of
Ba(OH)2:
1 liter Ba(OH)2
25.00 mL Ba(OH)2 x
= 0.02500 L Ba(OH)2
1000 mL Ba(OH)2
Finally, divide the moles of Ba(OH)2 by the
volume of Ba(OH)2:
Molarity [Ba(OH) 2 ] =
6.865 x 10 -3 moles Ba(OH) 2
= 0.2746 M Ba(OH) 2
0.02500 L Ba(OH) 2
11- 49
Acid-Base Titrations
Volume
of
Reactant
M
reactant
Moles
of
Reactant
mole
ratio
Moles
of
Product
M
product
Volume
of
Product
11-50
Colligative Properties
• A property that does not depend on the
identity of a solute in solution
• Vary only with the number of solute
particles present in a specific quantity of
solvent
• 4 colligative properties:
–
–
–
–
Osmotic pressure
Vapor pressure lowering
Boiling point elevation
Freezing point depression
11-51
17
Osmotic Pressure
•
Osmosis
– A process in which solvent
molecules diffuse through a
barrier that does not allow the
passage of solute particles
•
The barrier is called a
semipermeable membrane.
– A membrane that allows the
passage of some substances
but not others
•
Osmotic Pressure
– Pressure that can be exerted
on the solution to prevent
osmosis
11-52
Osmotic Pressure
Figure 11.28
11-53
Vapor Pressure Lowering
• Solutes come in 2 forms:
– Volatile
• Solutes that readily form a gas
– Nonvolatile
• Solutes that DO NOT readily form a gas
• Generally, the addition of a solute lowers the vapor
pressure of a solution when compared to the pure
solvent.
11-54
18
Vapor Pressure Lowering
11-55
Phase Diagram
11-56
Boiling Point Elevation
The addition of solute affects the boiling point
because it affects the vapor pressure.
• The boiling point is raised with the addition of
solute in comparison to the pure solvent.
• An equation that gives the increase in boiling
point:
∆Tb = Kbm
Where ∆Tb is the increase in temperature from the
pure solvent’s boiling point, Kb is the boiling
point constant, which is characteristic of a
particular solvent, and m is the molality (moles
of solute per kg of solution)
•
11-57
19
Freezing Point Depression
• The freezing point is lowered with the
addition of solute in comparison to the
pure solvent.
• An equation that gives the decrease in
freezing point:
∆Tf = Kfm
Where ∆Tf is the decrease in temperature
from the pure solvent’s freezing point, Kf
is the freezing point constant, which is
characteristic of a particular solvent, and
m is the molality (moles of solute per kg
of solution)
11-58
Practice – Freezing Point
Depression
• What is the change in freezing point
for a 1.5 m solution of sucrose in
water?
11-59
Practice Solutions – Freezing Point
Depression
• What is the change in freezing point
for a 1.5 m solution of sucrose in
water?
Kf (water) = -1.86°C/m
∆Tf = Kfm
∆Tf = -1.86°C * 1.5 m = -2.79°C
m
11-60
20
Colligative Properties and
Strong Electrolytes
•
•
•
•
Strong electrolytes dissociate most of the time
into their constituent ions.
Therefore, the number of particles (in this case
ions) increases with the number of ions.
Colligative properties are proportional to
number of particles in solution.
Example:
MgCl2(s) Mg2+(aq) + 2 Cl-(aq)
In this case the number of particles increases
to 3 particles. Therefore, we would multiply the
colligative property amount by 3.
11-61
Practice – Strong vs. Weak
Electrolytes
• Which of the following aqueous
solutions is expected to have the
lowest freezing point?
– 0.5 m CH3CH2OH
– 0.5 m Ca(NO3)2
– 0.5 m KBr
11-62
Practice Solutions – Strong vs.
Weak Electrolytes
•
Which of the following aqueous solutions is expected
to have the lowest freezing point?
–
–
–
0.5 m CH3CH2OH
0.5 m Ca(NO3)2
0.5 m KBr
Because the molalities are the same, we only
need to concentrate on the number of particles
generated by each solute. CH3CH2OH will not
dissociate into ions, Ca(NO3)2 dissociates into 3
ions, and KBr dissociates into 2 ions.
Therefore, the largest number of particles is
generated by Ca(NO3)2 and it will have the
lowest freezing point.
11-63
21